i've searched many webpages and yet can't find a specific and easy answer, so im writting here and hoping to get the answer. How do i solve this equation on Octave graphically?
There is the system:
system to solve
If the solutions are assumed to only be integers this can be a brief way to visually solve the equations by looking at the graphs. Note that the complementary case below the y-axis is don't covered here. Here both graphs are plotted on the same axes using the hold on function and is evaluated at integer intervals. Both functions are rearranged to be in terms of x. The mouse can then be used to click the intersections points. Hopefully, this serves as a starting point.
clf;
x = (-5: 1: 5);
y = sqrt(29 - x.^2);
plot(x,y,'.-');
hold on
x = (-5: 1: 5);
y = sqrt(9 + 4*x.^2);
plot(x,y,'.-');
Related
I have an equation:
b*cos(alpha) - a*sin(alpha) + b*cos(betta)-a*sin(betta) - b*cos(gamma) + a*sin(gamma) = 0
I want to create a 3D plot of this in Matlab with alpha vs betta vs gamma ( x - y - z ). I don't understand how to represent the equation so it could be plotted. How can I do this?
It is possible to assume that a = b = 1;
You need to understand what you have at hand. What does the equation show? Which form is it written? How does MATLAB plot different type of equations?
Fist, lets try to understand what type of function you have. It has 3 variables, but it equals to zero. A 3 variable equation generally defines a surface in 3D. In your case, this surface is described on its implicit form.
Now, if we look at the documentation of MATLAB, surfaces are generally plotted with surf, but surf needs 3 inputs (x,y,z) and you can not easily isolate your 3 variables.
Ah! but luckily, there is a thing called a search engine, that can gives us hints. Now that we know what kind of equation we have, we might as well use Google (or your favorite search engine) and type "implicit surface plot MATLAB", and that search will return a function called fimplicit3.
I think it seems to work:
I have solved a set of ordinary differential equations to achieve a solution as follows:
f = #(t,x) [(-B/m)*x(1);-g-(B/m)*x(2);x(1);x(2)];
[t,x_sol] = ode45(f,[0 5],[V_0*cos(alpha) V_0*sin(alpha) 0 0]);
and plotted one of the solutions as follows:
plot(t,x_sol(:,4));
All I need to do is to find the time where Y(t) becomes zero. But when I use t_f = interp1(x_sol(:,4),t,0) I get the first solution of 0. This might have a simple solution but I cant seem to get the other solution where Y becomes zero (which seems to be something around 4 seconds from the graph). This is actually a trajectory problem and all I simply need to do is to find the trajectory time.
Here is the given system I want to plot and obtain the vector field and the angles they make with the x axis. I want to find the index of a closed curve.
I know how to do this theoretically by choosing convenient points and see how the vector looks like at that point. Also I can always use
to compute the angles. However I am having trouble trying to code it. Please don't mark me down if the question is unclear. I am asking it the way I understand it. I am new to matlab. Can someone point me in the right direction please?
This is a pretty hard challenge for someone new to matlab, I would recommend taking on some smaller challenges first to get you used to matlab's conventions.
That said, Matlab is all about numerical solutions so, unless you want to go down the symbolic maths route (and in that case I would probably opt for Mathematica instead), your first task is to decide on the limits and granularity of your simulated space, then define them so you can apply your system of equations to it.
There are lots of ways of doing this - some more efficient - but for ease of understanding I propose this:
Define the axes individually first
xpts = -10:0.1:10;
ypts = -10:0.1:10;
tpts = 0:0.01:10;
The a:b:c syntax gives you the lower limit (a), the upper limit (c) and the spacing (b), so you'll get 201 points for the x. You could use the linspace notation if that suits you better, look it up by typing doc linspace into the matlab console.
Now you can create a grid of your coordinate points. You actually end up with three 3d matrices, one holding the x-coords of your space and the others holding the y and t. They look redundant, but it's worth it because you can use matrix operations on them.
[XX, YY, TT] = meshgrid(xpts, ypts, tpts);
From here on you can perform whatever operations you like on those matrices. So to compute x^2.y you could do
x2y = XX.^2 .* YY;
remembering that you'll get a 3d matrix out of it and all the slices in the third dimension (corresponding to t) will be the same.
Some notes
Matlab has a good builtin help system. You can type 'help functionname' to get a quick reminder in the console or 'doc functionname' to open the help browser for details and examples. They really are very good, they'll help enormously.
I used XX and YY because that's just my preference, but I avoid single-letter variable names as a general rule. You don't have to.
Matrix multiplication is the default so if you try to do XX*YY you won't get the answer you expect! To do element-wise multiplication use the .* operator instead. This will do a11 = b11*c11, a12 = b12*c12, ...
To raise each element of the matrix to a given power use .^rather than ^ for similar reasons. Likewise division.
You have to make sure your matrices are the correct size for your operations. To do elementwise operations on matrices they have to be the same size. To do matrix operations they have to follow the matrix rules on sizing, as will the output. You will find the size() function handy for debugging.
Plotting vector fields can be done with quiver. To plot the components separately you have more options: surf, contour and others. Look up the help docs and they will link to similar types. The plot family are mainly about lines so they aren't much help for fields without creative use of the markers, colours and alpha.
To plot the curve, or any other contour, you don't have to test the values of a matrix - it won't work well anyway because of the granularity - you can use the contour plot with specific contour values.
Solving systems of dynamic equations is completely possible, but you will be doing a numeric simulation and your results will again be subject to the granularity of your grid. If you have closed form solutions, like your phi expression, they may be easier to work with conceptually but harder to get working in matlab.
This kind of problem is tractable in matlab but it involves some non-basic uses which are pretty hard to follow until you've got your head round Matlab's syntax. I would advise to start with a 2d grid instead
[XX, YY] = meshgrid(xpts, ypts);
and compute some functions of that like x^2.y or x^2 - y^2. Get used to plotting them using quiver or plotting the coordinates separately in intensity maps or surfaces.
Say for example I have data which forms the parabolic curve y=x^2, and I want to read off the x value for a given y value. How do I go about doing this in MATLAB?
If it were a straight line, I could just use the equation of the line of best fit to calculate easily, however I can't do this with a curved line. If I can't find a solution, I'll solve for roots
Thanks in advance.
If all data are arrays (not analytical expressions), I usually do that finding minimal absolute error
x=some_array;
[~,ind]=min(abs(x.^2-y0))
Here y0 is a given y value
If your data are represented by a function, you can use fsolve:
function y = myfun(x)
y=x^2-y0
[x,fval] = fsolve(#myfun,x0,options)
For symbolic computations, one can use solve
syms x
solve(x^2 - y0)
Assuming your two curves are just two vectors of data, I would suggest you use Fast and Robust Curve Intersections from the File Exchange. See also these two similar questions: how to find intersection points when lines are created from an array and Finding where plots may cross with octave / matlab.
I would like to use a MATLAB function to find the minimum length between a point and a curve? The curve is described by a complicated function that is not quite smooth. So I hope to use an existing tool of matlab to compute this. Do you happen to know one?
When someone says "its complicated" the answer is always complicated too, since I never know exactly what you have. So I'll describe some basic ideas.
If the curve is a known nonlinear function, then use the symbolic toolbox to start with. For example, consider the function y=x^3-3*x+5, and the point (x0,y0) =(4,3) in the x,y plane.
Write down the square of the distance. Euclidean distance is easy to write.
(x - x0)^2 + (y - y0)^2 = (x - 4)^2 + (x^3 - 3*x + 5 - 3)^2
So, in MATLAB, I'll do this partly with the symbolic toolbox. The minimal distance must lie at a root of the first derivative.
sym x
distpoly = (x - 4)^2 + (x^3 - 3*x + 5 - 3)^2;
r = roots(diff(distpoly))
r =
-1.9126
-1.2035
1.4629
0.82664 + 0.55369i
0.82664 - 0.55369i
I'm not interested in the complex roots.
r(imag(r) ~= 0) = []
r =
-1.9126
-1.2035
1.4629
Which one is a minimzer of the distance squared?
subs(P,r(1))
ans =
35.5086
subs(P,r(2))
ans =
42.0327
subs(P,r(3))
ans =
6.9875
That is the square of the distance, here minimized by the last root in the list. Given that minimal location for x, of course we can find y by substitution into the expression for y(x)=x^3-3*x+5.
subs('x^3-3*x+5',r(3))
ans =
3.7419
So it is fairly easy if the curve can be written in a simple functional form as above. For a curve that is known only from a set of points in the plane, you can use my distance2curve utility. It can find the point on a space curve spline interpolant in n-dimensions that is closest to a given point.
For other curves, say an ellipse, the solution is perhaps most easily solved by converting to polar coordinates, where the ellipse is easily written in parametric form as a function of polar angle. Once that is done, write the distance as I did before, and then solve for a root of the derivative.
A difficult case to solve is where the function is described as not quite smooth. Is this noise or is it a non-differentiable curve? For example, a cubic spline is "not quite smooth" at some level. A piecewise linear function is even less smooth at the breaks. If you actually just have a set of data points that have a bit of noise in them, you must decide whether to smooth out the noise or not. Do you wish to essentially find the closest point on a smoothed approximation, or are you looking for the closest point on an interpolated curve?
For a list of data points, if your goal is to not do any smoothing, then a good choice is again my distance2curve utility, using linear interpolation. If you wanted to do the computation yourself, if you have enough data points then you could find a good approximation by simply choosing the closest data point itself, but that may be a poor approximation if your data is not very closely spaced.
If your problem does not lie in one of these classes, you can still often solve it using a variety of methods, but I'd need to know more specifics about the problem to be of more help.
There's two ways you could go about this.
The easy way that will work if your curve is reasonably smooth and you don't need too high precision is to evaluate your curve at a dense number of points and simply find the minimum distance:
t = (0:0.1:100)';
minDistance = sqrt( min( sum( bxsfun(#minus, [x(t),y(t)], yourPoint).^2,2)));
The harder way is to minimize a function of t (or x) that describes the distance
distance = #(t)sum( (yourPoint - [x(t),y(t)]).^2 );
%# you can use the minimum distance from above as a decent starting guess
tAtMin = fminsearch(distance,minDistance);
minDistanceFitte = distance(tAtMin);